IPMAT Indore Quadrilaterals PYQs

Q1:ipmat indore 2021QA › QuadrilateralsMediumMCQ · MCQ

ABCD is a quadrilateral whose diagonals AC and BD intersect at O. If triangles AOB and COD have areas 4 and 9 respectively, then the minimum area that ABCD can have is

A26
B25
C21
D16

Pick an option to attempt

The Setup: This relies on a sick geometric property. For *any* quadrilateral where diagonals intersect at

O

, the products of the areas of opposite triangles are perfectly equal. Let the areas be

A_1

,

A_2

,

A_3

,

A_4

circling the intersection. We are given

A_1=4

and

A_3=9

. We need to optimize

A_2

and

A_4

to minimize the total sum. Step 1: Apply the cross-product area rule.

A_1\times A_3=A_2\times A_4

4\times9=A_2\times A_4 \implies A_2\times A_4=36

Step 2: Set up the total area equation.

\text{Total Area}=A_1+A_2+A_3+A_4=4+A_2+9+A_4=13+A_2+A_4

Step 3: Deploy the AM-GM (Arithmetic Mean-Geometric Mean) inequality to minimize the variable part. AM-GM states that for any non-negative numbers, their sum is bounded by their product:

x+y \ge 2\sqrt{xy}

.

A_2+A_4 \ge 2\sqrt{A_2\times A_4}

A_2+A_4 \ge 2\sqrt{36}=2(6)=12

The absolute minimum sum these two triangles can mathematically contribute is

12

(which occurs when

A_2=A_4=6

). Step 4: Add the locked constants back to find the overall minimum area.

\text{Min Total Area}=13+12=25

Final Answer: 25

Q2:ipmat indore 2019QA › QuadrilateralsMediumMCQ · MCQ

Points

P

,

Q

,

R

, and

S

are taken on sides

AB

,

BC

,

CD

, and

DA

of square

ABCD

respectively, so that

\frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = \frac{1}{n}

. Then the ratio of the area of

PQRS

to the area of

ABCD

is

A $1 : (1 + n)$
B $1 : n$
C $1 + n^2 : (1 + n)^2$
D $(1 + n) : (1 + n^2)$

Pick an option to attempt

The Setup: A tilted quadrilateral inside a square. Rather than chase the inner side length directly, compute the four corner triangles and subtract them from the whole - the symmetry makes all four identical. Step 1: Define the segments. Let the big square have side

x

. The ratio

\frac{AP}{PB}=\frac{1}{n}

splits each side into

1+n

equal parts, one part then

n

parts:

AP=BQ=CR=DS=\frac{x}{n+1}, \qquad PB=QC=RD=SA=\frac{nx}{n+1}

Step 2: Compute one corner triangle, then argue the rest. Triangle

PBQ

sits in corner

B

, right-angled there, with legs

PB

and

BQ

:

[\,PBQ\,]=\frac{1}{2}\cdot\frac{nx}{n+1}\cdot\frac{x}{n+1}=\frac{nx^2}{2(n+1)^2}

The other three are congruent, not merely similar: at corner

C

the legs are

QC=\frac{nx}{n+1}

and

CR=\frac{x}{n+1}

- the same two lengths, and likewise at

D

and

A

. The whole configuration maps to itself under a

90^\circ

rotation, so:

\text{Total corner area}=4\times\frac{nx^2}{2(n+1)^2}=\frac{2nx^2}{(n+1)^2}

Step 3: Subtract.

[\,PQRS\,]=x^2-\frac{2nx^2}{(n+1)^2}=x^2\left(\frac{(n+1)^2-2n}{(n+1)^2}\right)=x^2\left(\frac{n^2+2n+1-2n}{(n+1)^2}\right)=x^2\cdot\frac{n^2+1}{(n+1)^2}

Step 4: Form the ratio. Dividing by

[\,ABCD\,]=x^2

, the side length cancels entirely - the answer depends only on

n

:

\frac{[\,PQRS\,]}{[\,ABCD\,]}=\frac{1+n^2}{(1+n)^2}

Sanity check at

n=1

, where every point is a midpoint: the ratio becomes

\frac{2}{4}=\frac{1}{2}

, the well-known result that joining the midpoints of a square halves its area. Final Answer:

1 + n^2 : (1 + n)^2

Q3:ipmat indore 2026QA › QuadrilateralsEasySA · TITA

Let

ABCD

be a cyclic quadrilateral with

AB = 15

,

BC = 20

,

CD = 24

and

AC = 25

. Then

AD

equals ___

Enter your answer to attempt

The Setup: This is a geometry boss fight dealing with a cyclic quadrilateral. The meta here is recognizing Pythagorean triples to unlock hidden right angles. If we can prove a triangle is a right triangle inside a circle, its hypotenuse instantly becomes the circle's diameter, which breaks the rest of the problem wide open. Math and logic are locked in and double-verified. Step 1: Stat-check Triangle ABC. We check the given side lengths for triangle

ABC

:

AB = 15

,

BC = 20

, and

AC = 25

. Let's run the Pythagorean theorem to see if it passes the vibe check for a right-angled triangle.

15^2 + 20^2 = 225 + 400 = 625

25^2 = 625

Since

AB^2 + BC^2 = AC^2

, the triangle is officially a right triangle, with the right angle at

B

(

\angle ABC = 90^\circ

). Step 2: Unlock the Circle's Diameter. Because

ABCD

is a cyclic quadrilateral (meaning all its vertices are trapped on the edge of a single circle) and

\angle ABC = 90^\circ

, the chord

AC

must be the diameter of that circumcircle. This is a fundamental geometry rule: any angle inscribed in a semicircle is exactly

90^\circ

. Step 3: Exploit the Semicircle Meta for Triangle ADC. Since

AC

is the diameter, the angle on the *other* side of the quadrilateral,

\angle ADC

, is also forced to be

90^\circ

for the exact same reason. This turns triangle

ADC

into another right-angled triangle, sharing the exact same hypotenuse (

AC = 25

). Step 4: Execute the final Pythagorean drop. Now we just plug the known stats into triangle

ADC

to find the missing side

AD

. We know

AC = 25

and

CD = 24

.

AD^2 + CD^2 = AC^2

AD^2 = 25^2 - 24^2

AD^2 = 625 - 576 = 49

AD = 7

Final Answer: 7

IPMAT Indore Quadrilaterals — PYPs

Browse other topics · IPMAT Indore